1 Bad Mirada

but 8-10 psi is running pretty steep on a stock 10.3 to 1 engine...

also, those HP numbers are whp, from what vehicle? different vehicles are more or less efficient as pertaining to rwhp.

standard air pressure is 13.8, i believe. so every 13.8 psi added by forced induction will, in theory, double the horsepower, but this does not take into account any type of parasitic loss or basic drivetrain loss. so, if you use the assumption (which i reiterate that is not by any means perfect), that x=14psi, and y=psi produced by the blower without loss of any kind, and z=stock bhp, you would get this, solving for B

(y/x+1)*z=B

say youre making 5psi, on a 100hp motor, then you would have this:
(5/14+1)*100=135.7bhp

so, running 10psi on a 425hp motor, would give you this.
(10/14+1)*425=728.57bhp

now, figure that the car has a 25% drivetrain loss...so 728.57*.75=546.43whp, minus the power eaten by the supercharger (parasitic loss).

i dont recall roughly how much power the blower itself eats, but i seem to recall that someone did a test with a C5 corvette with a 325hp LS1 (i think), and at 3psi, the supercharged car was making the same amount of RWHP as it did prior to the blower being fitted up. so, assuming that the first three psi of boost is making the hp needed to spin the supercharger, you would be running 7 pounds of "useful" boost, so
[(7/14+1)*425]*.75=478rwhp.

mind you, this is all very conceptual, and there are SO many variables that without an engine dyno, it is little more than an educated guess. i higher hp engine will make more boost with the first 3psi of boost that the corvette used to "turn" the blower...so it woudl take less effort by the motor, thus freeing up more WHP.